Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $x = \dfrac{-8k - 24}{-2k + 10} \div \dfrac{-2k + 4}{k^2 - 7k + 10} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{-8k - 24}{-2k + 10} \times \dfrac{k^2 - 7k + 10}{-2k + 4} $ First factor the quadratic. $x = \dfrac{-8k - 24}{-2k + 10} \times \dfrac{(k - 5)(k - 2)}{-2k + 4} $ Then factor out any other terms. $x = \dfrac{-8(k + 3)}{-2(k - 5)} \times \dfrac{(k - 5)(k - 2)}{-2(k - 2)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ -8(k + 3) \times (k - 5)(k - 2) } { -2(k - 5) \times -2(k - 2) } $ $x = \dfrac{ -8(k + 3)(k - 5)(k - 2)}{ 4(k - 5)(k - 2)} $ Notice that $(k - 2)$ and $(k - 5)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ -8(k + 3)\cancel{(k - 5)}(k - 2)}{ 4\cancel{(k - 5)}(k - 2)} $ We are dividing by $k - 5$ , so $k - 5 \neq 0$ Therefore, $k \neq 5$ $x = \dfrac{ -8(k + 3)\cancel{(k - 5)}\cancel{(k - 2)}}{ 4\cancel{(k - 5)}\cancel{(k - 2)}} $ We are dividing by $k - 2$ , so $k - 2 \neq 0$ Therefore, $k \neq 2$ $x = \dfrac{-8(k + 3)}{4} $ $x = -2(k + 3) ; \space k \neq 5 ; \space k \neq 2 $